An electron accelerated from rest through a voltage of 330 V

An electron accelerated from rest through a voltage of 330 V enters a region of constant magnetic field. If the electron follows a circular path with a radius of 16 cm, what is the magnitude of the magnetic field? It wants the answer in mT not T. I think I\'m close but it says I\'m off by a multiple of ten, I\'m just so stumped. I\'m setting it up so far with B=sqrt((2meV)/(r^2*e)) with charge of an electron e=1.6*10^-19, and mass of electron me=9.1*10^-31, I converted 16cm to 0.16m. The answer I got was 3.8*10^-4 which I guess is in T not mT I tried converting it but I guess that wasn\'t right either. Where am I going wrong?

Solution

q E = 1/2 m v²

(1.6 * 10^-19) * 330 = 1/2 * (9.11 * 10^-31) * v²

v² = [(1.6 * 10^-19) * 330] / [1/2 * (9.11 * 10^-31)]

v = 1.08 * 10⁷ m/s

r = m v / q B

B = m v / q r = (9.11 * 10^-31 * 1.08 * 10⁷) / (1.6 * 10^-19 * 0.16)

B = 0.000383 T = 3.8 * 10^-4 T

= 0.383 mT